5

Please consider the following list:

There is a $(K_n)_{n\in\mathbb N}\subseteq E$ with
\begin{itemize}
    \item $K_n$ is compact;
    \item $K_n\subseteq\overset\circ K_{n+1}$; and
    \item $d(\partial K_n,\partial K_{n+1})>0$
\end{itemize}
for all $n\in\mathbb N$.

This doesn't render well:

render

There's too much vertical empty space between the first and the second item (due to the circle above K). I would like to set the vertical spacing between these two items to 0pt, but I only know how I can set the vertical spacing between all items or between all items starting from a particular item.

So, how can we do this? Or would you recommend an other solution? Maybe it would be better to use only symbols which don't extend vertically inside a list. How is this usually done? Do you use $K^\circ$ instead? Or $\operatorname{int} K$? I don't like the latter and my problem with $K^\circ$ is that it's not consistent with other occurences in my document where I've used $\overset\circ K$.

  • Adjust vertical space manually with \[-1.5ex] at the end of each \item. Change 1.5 to get desired results. – jak123 Jun 23 at 12:14
  • @jak123 Why at the end of each item? – 0xbadf00d Jun 23 at 12:27
  • Each item you would like to adjust. And its \\ not \ as I wrote before. – jak123 Jun 23 at 16:12
7

You can use \smash{\overset\circ K_{n+1}} to hide the height of that element.

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\newtheorem{theorem}{Theorem}[section]
\begin{document}

\begin{theorem}
There is a $(K_n)_{n\in\mathbb N}\subseteq E$ with
\begin{itemize}
    \item $K_n$ is compact;
    \item $K_n\subseteq\smash{\overset\circ K_{n+1}}$; and
    \item $d(\partial K_n,\partial K_{n+1})>0$
\end{itemize}
for all $n\in\mathbb N$.
\end{theorem}

\end{document} 

enter image description here

  • Perfect, thank you very much. – 0xbadf00d Jun 23 at 12:57
4

You don't have this problem if you use the \mathring accent:

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\newtheorem{theorem}{Theorem}[section]

\begin{document}

\begin{theorem}
There is a $(K_n)_{n\in\mathbb N}\subseteq E$ with
\begin{itemize}
    \item $K_n$ is compact;
    \item $K_n\subseteq \ring{K}_{n+1} $; and
    \item $d(\partial K_n,\partial K_{n+1})>0$
\end{itemize}
for all $n\in\mathbb N$.
\end{theorem}

\end{document} 

enter image description here

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.